如何使用scikit线性回归模型同时解决几个独立的时间序列

我尝试使用sklearn线性回归模型同时预测多个独立时间序列,但我似乎无法做到正确.

我的数据组织如下：Xn是一个矩阵,其中每行包含4个观测值的预测窗口,yn是Xn每行的目标值.

import numpy as np

# training data
X1=np.array([[-0.31994,-0.32648,-0.33264,-0.33844],[-0.32648,-0.33264,-0.33844,-0.34393],[-0.33264,-0.33844,-0.34393,-0.34913],[-0.33844,-0.34393,-0.34913,-0.35406],[-0.34393,-0.34913,-.35406,-0.35873],[-0.34913,-0.35406,-0.35873,-0.36318],[-0.35406,-0.35873,-0.36318,-0.36741],[-0.35873,-0.36318,-0.36741,-0.37144],[-0.36318,-0.36741,-0.37144,-0.37529],[-0.36741,-.37144,-0.37529,-0.37896],[-0.37144,-0.37529,-0.37896,-0.38069],[-0.37529,-0.37896,-0.38069,-0.38214],[-0.37896,-0.38069,-0.38214,-0.38349],[-0.38069,-0.38214,-0.38349,-0.38475],[-.38214,-0.38349,-0.38475,-0.38593],[-0.38349,-0.38475,-0.38593,-0.38887]])
X2=np.array([[-0.39265,-0.3929,-0.39326,-0.39361],[-0.3929,-0.39326,-0.39361,-0.3931],[-0.39326,-0.39361,-0.3931,-0.39265],[-0.39361,-0.3931,-0.39265,-0.39226],[-0.3931,-0.39265,-0.39226,-0.39193],[-0.39265,-0.39226,-0.39193,-0.39165],[-0.39226,-0.39193,-0.39165,-0.39143],[-0.39193,-0.39165,-0.39143,-0.39127],[-0.39165,-0.39143,-0.39127,-0.39116],[-0.39143,-0.39127,-0.39116,-0.39051],[-0.39127,-0.39116,-0.39051,-0.3893],[-0.39116,-0.39051,-0.3893,-0.39163],[-0.39051,-0.3893,-0.39163,-0.39407],[-0.3893,-0.39163,-0.39407,-0.39662],[-0.39163,-0.39407,-0.39662,-0.39929],[-0.39407,-0.39662,-0.39929,-0.4021]])

# target values
y1=np.array([-0.34393,-0.34913,-0.35406,-0.35873,-0.36318,-0.36741,-0.37144,-0.37529,-0.37896,-0.38069,-0.38214,-0.38349,-0.38475,-0.38593,-0.38887,-0.39184])
y2=np.array([-0.3931,-0.39265,-0.39226,-0.39193,-0.39165,-0.39143,-0.39127,-0.39116,-0.39051,-0.3893,-0.39163,-0.39407,-0.39662,-0.39929,-0.4021,-0.40506])

单个时间序列的正常过程按预期工作,如下所示：

from sklearn.linear_model import LinearRegression

# train the 1st half, predict the 2nd half
half = len(y1)/2 # or y2 as they have the same length
LR = LinearRegression()
LR.fit(X1[:half], y1[:half])
pred = LR.predict(X1[half:])
r_2 = LR.score(X1[half:],y1[half:])

但是如何将线性回归模型同时应用于多个独立的时间序列？
我尝试了以下方法：

y_stack = np.vstack((y1[None],y2[None]))
X_stack = np.vstack((X1[None],X2[None]))

print 'y1 shape:',y1.shape, 'X1 shape:',X1.shape
print 'y_stack shape:',y_stack.shape, 'X_stack:',X_stack.shape
y1 shape: (16,) X1 shape: (16, 4)
y_stack shape: (2, 16) X_stack: (2, 16, 4)

但是线性模型的拟合失败如下：

LR.fit(X_stack[:,half:],y_stack[:,half:])

说明维度数量高于预期：

C:\Python27\lib\site-packages\sklearn\utils\validation.pyc in check_array(array, accept_sparse, dtype, order, copy, force_all_finite, ensure_2d, allow_nd, ensure_min_samples, ensure_min_features, warn_on_dtype, estimator)
        394         if not allow_nd and array.ndim >= 3:
        395             raise ValueError("Found array with dim %d. %s expected <= 2."
    --> 396                              % (array.ndim, estimator_name))
        397         if force_all_finite:
        398             _assert_all_finite(array)

    ValueError: Found array with dim 3. Estimator expected <= 2.

任何建议或提示都非常感谢.

UPDATE

我可以使用for循环,但由于n实际上是10000或更多的顺序,我希望找到包含数组操作的解决方案,因为这些是numpy,scipy和hopelely sklearn的显式功能

解决方法:

@ali_m我不认为这是一个重复的问题,但它们部分相关.
当然,使用类似于sklearn的线性回归模型可以同时应用和预测时间序列：

我创建了一个新类LinearRegression_Multi：

class LinearRegression_Multi:
    def stacked_lstsq(self, L, b, rcond=1e-10):
        """
        Solve L x = b, via SVD least squares cutting of small singular values
        L is an array of shape (..., M, N) and b of shape (..., M).
        Returns x of shape (..., N)
        """
        u, s, v = np.linalg.svd(L, full_matrices=False)
        s_max = s.max(axis=-1, keepdims=True)
        s_min = rcond*s_max
        inv_s = np.zeros_like(s)
        inv_s[s >= s_min] = 1/s[s>=s_min]
        x = np.einsum('...ji,...j->...i', v,
                      inv_s * np.einsum('...ji,...j->...i', u, b.conj()))
        return np.conj(x, x)    

    def center_data(self, X, y):
        """ Centers data to have mean zero along axis 0. 
        """
        # center X        
        X_mean = np.average(X,axis=1)
        X_std = np.ones(X.shape[0::2])
        X = X - X_mean[:,None,:] 
        # center y
        y_mean = np.average(y,axis=1)
        y = y - y_mean[:,None]
        return X, y, X_mean, y_mean, X_std

    def set_intercept(self, X_mean, y_mean, X_std):
        """ Calculate the intercept_
        """
        self.coef_ = self.coef_ / X_std # not really necessary
        self.intercept_ = y_mean - np.einsum('ij,ij->i',X_mean,self.coef_)

    def scores(self, y_pred, y_true ):
        """ 
        The coefficient R^2 is defined as (1 - u/v), where u is the regression
        sum of squares ((y_true - y_pred) ** 2).sum() and v is the residual
        sum of squares ((y_true - y_true.mean()) ** 2).sum().        
        """        
        u = ((y_true - y_pred) ** 2).sum(axis=-1)
        v = ((y_true - y_true.mean(axis=-1)[None].T) ** 2).sum(axis=-1)
        r_2 = 1 - u/v
        return r_2

    def fit(self,X, y):
        """ Fit linear model.        
        """        
        # get coefficients by applying linear regression on stack
        X_, y, X_mean, y_mean, X_std = self.center_data(X, y)
        self.coef_ = self.stacked_lstsq(X_, y)
        self.set_intercept(X_mean, y_mean, X_std)

    def predict(self, X):
        """Predict using the linear model
        """
        return np.einsum('ijx,ix->ij',X,self.coef_) + self.intercept_[None].T

可以使用与问题中相同的声明变量,如下所示：

LR_Multi = LinearRegression_Multi()
LR_Multi.fit(X_stack[:,:half], y_stack[:,:half])
y_stack_pred = LR_Multi.predict(X_stack[:,half:])
R2 = LR_Multi.scores(y_stack_pred, y_stack[:,half:])

多个时间序列的R ^ 2是：

array([ 0.91262442,  0.67247516])

这确实类似于标准sklearn线性回归的预测方法：

from sklearn.linear_model import LinearRegression

LR = LinearRegression()
LR.fit(X1[:half], y1[:half])
R2_1 = LR.score(X1[half:],y1[half:])

LR.fit(X2[:half], y2[:half])
R2_2 = LR.score(X2[half:],y2[half:])
print R2_1, R2_2
0.912624422097 0.67247516054

标签：python,numpy,scikit-learn,linear-regression
来源： https://codeday.me/bug/20190702/1356445.html